While doing research in computer system, I came across the following summation:
$$S_n = \sum_{k=0}^{n} {n \choose k} k! = \sum_{k=0}^{n} \frac{n!}{(n-k)!}$$
where both $n$ and $k$ are integers.
$S_n$ ...

A teacher gave this as a homework question, and I have tried but haven't been able to arrive at a solution.
$\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$
Could someone prove it, or at least ...

Can we find $$ \sum_{k=0}^{n} \sqrt{\binom{n}{k}} \quad$$
This problem asked me my friend about a year ago, but I didn't know how to attack problem. Now, I am interesting in solution. Any suggestion?
...

The other day a friend of mine showed me this sum: $\sum_{k=0}^n\binom{3n}{3k}$. To find the explicit formula I plugged it into mathematica and got $\frac{8^n+2(-1)^n}{3}$. I am curious as to how one ...

Just wondering about this:
Is it true that there are no prime numbers in Pascal's triangle, with the exception of $\binom{n}{1}$ and $\binom{n}{n-1}$?
From the first 18 lines it appears that this is ...

We can view the binomial coefficient $\binom{x}{k}$ has a polynomial in $x$ with degree $k$. So taking some $f\in\mathbb{Q}[x]$, why is $f(n)\in\mathbb{Z}$ for all $n\in\mathbb{Z}$, precisely when the ...

Show that ${-n \choose i} = (-1)^i{n+i-1 \choose i} $. This is a homework exercise I have to make and I just cant get started on it. The problem lies with the $-n$. Using the definition I get:
$${-n ...

$\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,\ldots,\binom{2n}{2n-1}\right)$
i want to know what is specialty of such a series.I am not able to generalize the problem solution.Is there any rule for ...

I'd like to get a hint to prove the following identity:
$$\tag{1}\sum_{\nu}(-1)^{\nu}\displaystyle \binom{a}{\nu}\binom{n-\nu}{r}=\binom{n-a}{n-r} .$$
The original statement reads "By specialization ...

Some background.
I was asked to find an arithmetic function $f$ such that $f*f=\mathbf 1$ where $\mathbf 1$ is the constant function 1 and $*$ denotes Dirichlet convolution. I was able to prove that ...

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$
I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively ...

Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ ...